Rational operators (a {t} b); a,b > e solved

[JmsNxn]

Hi James,

\( \hspace{48} \begin{eqnarray}
a &+& b &=& a + b \\
a &+_{\tiny -1} & b &=& \log(e^a + e^b) \\
a &+_{\tiny -2} & b &=& \log(\log(e^e^a + e^e^b)) \\
\vspace8 \end{eqnarray} \)

Yes this is right:
\( a\,\,\bigtriangleup_{-1}^e\,\,b = \ln(e^a+e^b) \)

I did a lot of investigation into this operator (well, to the best that I could).

and \( L^{\tiny o h}(3) \) for the h-fold iterated log( 3) then Mike's limit can be expressed
\( \begin{eqnarray}
t_1 &=& L^{\tiny o 1}(3) \\
t_2 &=& L^{\tiny o 1}(3) &+& L^{\tiny o 2}(3) \\
t_3 &=& L^{\tiny o 1}(3) &+& L^{\tiny o 2}(3) &+_{\tiny -1}& L^{\tiny o 3}(3) \\

t_4 &=& L^{\tiny o 1}(3) &+& L^{\tiny o 2}(3) &+_{\tiny -1}& L^{\tiny o 3}(3) &+_{\tiny -2}& L^{\tiny o 4}(3) \\
\vspace8 & &\\
...& &... \\
\vspace8 & &\\
t_{n\to \infty} &\to & \text{constant} \\
\end{eqnarray} \)
where the operator-precedence is lower the more negative the index at the plus is (so we evaluate it from the left).

First question: is this in fact an application of your "rational operator"?

Well it appears to be. I'm floored, I'm terrible at finding applications.

And if it is so, then second question: does this help to evaluate this to higher depth of iteration than we can do it when we try it just by log and exp alone (we can do it to iteration 4 or 5 at max I think) ?

Well, not so far since the calculations involved in lower order operators rely on iterations of exp. However, I investigated in seeing if \( f(x) = a\,\,\bigtriangleup_{-1}^e\,\,x \) was analytic (which should help in calculations). But I only made it to the sixth or seventh derivative before I realized I wasn't going to recognize the pattern. The thread's here http://www.mymathforum.com/viewtopic.php?f=23&t=20993 .
I assume it would be analytic, (the function looks analytic when graphed, if that's any argument). Also, I think that if \( a\,\,\bigtriangleup_{-1}^e\,\,x \) is analytic, then \( a\,\,\bigtriangleup_{-2}\,\,x \) is probably analytic, since it's basically the same function with just a faster convergence to y=x and a higher starting point at negative infinity.